Answer
$$\frac{2\sqrt{6}-\sqrt{3}}{3\sqrt{3}+\sqrt{6}}=\sqrt{2}-1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{6}-\sqrt{3}}{3\sqrt{3}+\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{6}-\sqrt{3}}{3\sqrt{3}+\sqrt{6}}\frac{3\sqrt{3}-\sqrt{6}}{3\sqrt{3}-\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18\sqrt{2}-12-9+3\sqrt{2}}{27-9\sqrt{2}+9\sqrt{2}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{21\sqrt{2}-21}{21} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{2}-1}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{2}-1\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{3}- \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{6}- \sqrt{3}\right) } \cdot \left( 3 \sqrt{3}- \sqrt{6}\right) = \color{blue}{ 2 \sqrt{6}} \cdot 3 \sqrt{3}+\color{blue}{ 2 \sqrt{6}} \cdot- \sqrt{6}\color{blue}{- \sqrt{3}} \cdot 3 \sqrt{3}\color{blue}{- \sqrt{3}} \cdot- \sqrt{6} = \\ = 18 \sqrt{2}-12-9 + 3 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{3} + \sqrt{6}\right) } \cdot \left( 3 \sqrt{3}- \sqrt{6}\right) = \color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot- \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot 3 \sqrt{3}+\color{blue}{ \sqrt{6}} \cdot- \sqrt{6} = \\ = 27- 9 \sqrt{2} + 9 \sqrt{2}-6 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 21. |
| ⑤ | Remove 1 from denominator. |
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